Merge branch 'master' into master

Author: ironartisan

problems/0001.两数之和.md

@@ -224,6 +224,31 @@ func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
 }
 ```
 
+PHP:
+```php
+class Solution {
+    /**
+     * @param Integer[] $nums
+     * @param Integer $target
+     * @return Integer[]
+     */
+    function twoSum($nums, $target) {
+        if (count($nums) == 0) {
+            return [];
+        }
+        $table = [];
+        for ($i = 0; $i < count($nums); $i++) {
+            $temp = $target - $nums[$i];
+            if (isset($table[$temp])) {
+                return [$table[$temp], $i];
+            }
+            $table[$nums[$i]] = $i;
+        }
+        return [];
+    }
+}
+```
+
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)

problems/0015.三数之和.md

@@ -393,6 +393,46 @@ function threeSum(array $nums): array
 }
 ```
 
+PHP:
+```php
+class Solution {
+    /**
+     * @param Integer[] $nums
+     * @return Integer[][]
+     */
+    function threeSum($nums) {
+        $res = [];
+        sort($nums);
+        for ($i = 0; $i < count($nums); $i++) {
+            if ($nums[$i] > 0) {
+                return $res;
+            }
+            if ($i > 0 && $nums[$i] == $nums[$i - 1]) {
+                continue;
+            }
+            $left = $i + 1;
+            $right = count($nums) - 1;
+            while ($left < $right) {
+                $sum = $nums[$i] + $nums[$left] + $nums[$right];
+                if ($sum < 0) {
+                    $left++;
+                }
+                else if ($sum > 0) {
+                    $right--;
+                }
+                else {
+                    $res[] = [$nums[$i], $nums[$left], $nums[$right]];
+                    while ($left < $right && $nums[$left] == $nums[$left + 1]) $left++;
+                    while ($left < $right && $nums[$right] == $nums[$right - 1]) $right--;
+                    $left++;
+                    $right--;
+                }
+            }
+        }
+        return $res;
+    }
+}
+```
 
 
 -----------------------

problems/0018.四数之和.md

@@ -310,6 +310,49 @@ var fourSum = function(nums, target) {
 };
 ```
 
+PHP:
+```php
+class Solution {
+    /**
+     * @param Integer[] $nums
+     * @param Integer $target
+     * @return Integer[][]
+     */
+    function fourSum($nums, $target) {
+        $res = [];
+        sort($nums);
+        for ($i = 0; $i < count($nums); $i++) {
+            if ($i > 0 && $nums[$i] == $nums[$i - 1]) {
+                continue;
+            }
+            for ($j = $i + 1; $j < count($nums); $j++) {
+                if ($j > $i + 1 && $nums[$j] == $nums[$j - 1]) {
+                    continue;
+                }
+                $left = $j + 1;
+                $right = count($nums) - 1;
+                while ($left < $right) {
+                    $sum = $nums[$i] + $nums[$j] + $nums[$left] + $nums[$right];
+                    if ($sum < $target) {
+                        $left++;
+                    }
+                    else if ($sum > $target) {
+                        $right--;
+                    }
+                    else {
+                        $res[] = [$nums[$i], $nums[$j], $nums[$left], $nums[$right]];
+                        while ($left < $right && $nums[$left] == $nums[$left+1]) $left++;
+                        while ($left < $right && $nums[$right] == $nums[$right-1]) $right--;
+                        $left++;
+                        $right--;
+                    }
+                }
+            }
+        }
+        return $res;
+    }
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0077.组合优化.md

@@ -242,8 +242,59 @@ var combine = function(n, k) {
 };
 ```
 
+C:
+```c
+int* path;
+int pathTop;
+int** ans;
+int ansTop;
+
+void backtracking(int n, int k,int startIndex) {
+    //当path中元素个数为k个时，我们需要将path数组放入ans二维数组中
+    if(pathTop == k) {
+        //path数组为我们动态申请，若直接将其地址放入二维数组，path数组中的值会随着我们回溯而逐渐变化
+        //因此创建新的数组存储path中的值
+        int* temp = (int*)malloc(sizeof(int) * k);
+        int i;
+        for(i = 0; i < k; i++) {
+            temp[i] = path[i];
+        }
+        ans[ansTop++] = temp;
+        return ;
+    }
+
+    int j;
+    for(j = startIndex; j <= n- (k - pathTop) + 1;j++) {
+        //将当前结点放入path数组
+        path[pathTop++] = j;
+        //进行递归
+        backtracking(n, k, j + 1);
+        //进行回溯，将数组最上层结点弹出
+        pathTop--;
+    }
+}
 
+int** combine(int n, int k, int* returnSize, int** returnColumnSizes){
+    //path数组存储符合条件的结果
+    path = (int*)malloc(sizeof(int) * k);
+    //ans二维数组存储符合条件的结果数组的集合。（数组足够大，避免极端情况）
+    ans = (int**)malloc(sizeof(int*) * 10000);
+    pathTop = ansTop = 0;
 
+    //回溯算法
+    backtracking(n, k, 1);
+    //最后的返回大小为ans数组大小
+    *returnSize = ansTop;
+    //returnColumnSizes数组存储ans二维数组对应下标中一维数组的长度（都为k）
+    *returnColumnSizes = (int*)malloc(sizeof(int) *(*returnSize));
+    int i;
+    for(i = 0; i < *returnSize; i++) {
+        (*returnColumnSizes)[i] = k;
+    }
+    //返回ans二维数组
+    return ans;
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)